A 6.0 -pF capacitor is needed to construct a circuit. The only capacitors available are rated as \(9.0 \mathrm{pF} .\) How can a combination of three 9.0 -pF capacitors be assembled so that the equivalent capacitance of the combination is \(6.0 \mathrm{pF} ?\)

: We will be working with two basic formulas for capacitors connected in series and parallel. For capacitors connected in parallel, the equivalent capacitance is given by the sum of the individual capacitances: \(C_\text{eq(parallel)} = C_1 + C_2 + C_3\). For capacitors connected in series, the reciprocal of the equivalent capacitance is given by the sum of the reciprocals of the individual capacitances: \(\frac{1}{C_\text{eq(series)}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}\).

: Let's try different configurations of capacitors and see if we can get the equivalent capacitance of 6.0 pF. - All three capacitors in parallel: \(C_\text{eq(parallel)} = C_1 + C_2 + C_3 = 9.0 + 9.0 + 9.0 = 27.0 \, \mathrm{pF}\) This configuration is not suitable as the equivalent capacitance is much higher than the needed 6.0 pF. - All three capacitors in series: \(\frac{1}{C_\text{eq(series)}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} = \frac{1}{9.0} + \frac{1}{9.0} + \frac{1}{9.0} = \frac{3}{9.0}\) Now, we found the reciprocal of the equivalent capacitance, so let's calculate the equivalent capacitance. \(C_\text{eq(series)} = \frac{1}{\frac{3}{9.0}}= 3.0 \, \mathrm{pF}\) This configuration is not suitable either, as the equivalent capacitance is lower than the desired 6.0 pF. - Two capacitors in parallel and one in series: First, we find the equivalent capacitance of the two capacitors in parallel: \(C_\text{eq(parallel)}= C_1 + C_2 = 9.0 + 9.0 = 18.0 \, \mathrm{pF}\) Next, we find the equivalent capacitance using one capacitor from the first step (C1 and C2 in parallel) and the remaining capacitor (C3) in series: \(\frac{1}{C_\text{eq(series)}} = \frac{1}{C_\text{eq(parallel)}} + \frac{1}{C_3} = \frac{1}{18.0} + \frac{1}{9.0}\) Calculate the equivalent capacitance, \(C_\text{eq(series)}= \frac{1}{\frac{1}{18.0} + \frac{1}{9.0}} = 6.0 \, \mathrm{pF}\) We have found the correct configuration! Two capacitors in parallel followed by one capacitor in series give an equivalent capacitance of 6.0 pF.